Foundations, part 11

$f(x)=x^{11} + 2 x^9 + 4x^7 + 8x^5 +16x^3 +32x - 64 \\ => f'(x)=11x^{10}+18x^8+28x^6+40 x^4+48 x^2+32$

As $f'(x)$ contains all even powers with positive coefficients then for all $x$ $f'(x)$ will be positive.

So, $f(x)$ is an increasing function and also $f(0)=-64$ .

Then the curve will cut the X-axis only once.Thus it has only one positive real root and $10$ complex roots.

Relevant wiki: Descartes' Rule of Signs

Let $f(x)=x^{11} + 2 x^9 + 4x^7 + 8x^5 +16x^3 +32x - 64$

then $f(-x)=-x^{11} - 2 x^9 - 4x^7 - 8x^5 - 16x^3 - 32x - 64$

By Descartes' rule of signs, the number of sign changes in $f(x)$ is $1$ , so there is $1$ positive root. And $f(-x)$ have no sign change, so there is no negative root. Hence only $1$ real root.

Since degree of equation is $11$ , hence number of non-real roots are $11 - 1 = \boxed{10}$

Derivative is positive, hence strictly increasing function which implies atmost one real root. At x=0, function value is negative and as x approaches to positive infinity, function tends to positive infinity. Hence, exactly one real (positive) root which means 10 complex roots.